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Unformatted text preview: EXAM #3
PHYS 126
19 APRIL 2007 Name: (please print legibly) I
Student ID #2 MW; Part Score Short answer Problem #1 yo [email protected] 2”; Tm] ° Problem #3
Total 2 Honor Code Exams must be completed within the allotted time and with
no outside assistance. Instructors will periodically enter the classroom, and
you are permitted to ask questions of them. However, remember that your
instructor is limited in the amount of information he can give. There should be
no communication with any other persons (verbal, electronic, etc.) other than
the instructor. The use of calculators on the exam is permitted, but is limited
to numerical calculation. Any other uses (unit conversion, formulas, graphing,
symbolic manipulation, etc.) are not allowed. Please sign the honor code below as conﬁrmation that you have complied
with this policy: “On my honor, I have neither given nor received any unauthorized
aid on this exam.” :1 lCl M! m (signature) date) PHYS 126 Exam #3 19 April 2007 Short Answer (40 points)
Each of the ten questions below is worth 4 points. 1. A straight wire is producing a steady current to the right.
W I XXXX
G? In which direction is current induced in the loop of wire? (a) clockwise
@counterclockwise (c) No current is induced in the loop. The current in the straight wire is shut off over a short time span. In
which direction does the loop feel a force during this time span? (a) up
(b) down @ the loop does not experience a force. 2. The circuit shown below is broken along the bottom wirev‘between points
1 and 2. 10v
1 AV :iKv
109 «7
0, What is the potential difference AV21 2 V2 — V1 between points 1 and 2?
(a) +10 V ((33+5V mo WJW/ﬁg W 0 V (d) —5v émtch bmm ‘ (e) —10 V PHYS 126
Exam #3 19 April 2007 3. A bar magnet falls southﬁrst through a horizontal loop of wire. What is the direction of induced current ﬂow in the loop of wire as the
magnet passes entirely through the loop? (d) 2 then 1 (e) No current is induced. 4, Consider the current carrying wire and the moving charge in the ﬁgure
below. Draw and label the magnetic force acting on the~charge at the
instant shown in the ﬁgure. Use an arrow for forces in the plane, a ® or
Q for forces perpendicular to the plane, and 6 for no force. Tm]
9———>i3 5. You have two choices in constructing a solenoid. You can use a standard
copper wire, or you can use a gold wire that is half as thick but can carry
twice the current. In either case you will wrap the solenoid so that the
coils are as close together as possible and make it such that the length is
much longer than the diameter of the solenoid, Which will maximize the magnetic ﬁeld in the solenoid? (a) The copper wire. @) The gold wire. (c) They create the same ﬁeld strength. limes/V} Exam #3 PHYS 126
19 April 2007 6. Charge ﬂows through a light bulb. Suppose a wire is connected across the bulb as shown. ’ @
4)
When the Wire is connected, all the charge ﬂows through the wire. (b) half the charge flows through the wire, the other half continues through
the bulb. (c) all the charge continues to ﬂow through the bulb. ((1) none of the above . In an experiment an electron and positron (same mass as an electron but opposite charge) are produced from the same point in space. Their
subsequent behavior in a magnetic ﬁeld is described below. X X X X X
X X X X
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5 \
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X 1 X \\><’/ x
x X X X X If the magnetic ﬁeld points into the page, which is the positron? (a) 1, the leftmost and solid path.
@ 2, the rightmost and dashed path. At the time of their creation, which is traveling fastest? 1, the leftmost and solid path.
(b) 2, the rightmost and dashed path. v..— r, PHYS 126 Exam #3 19 April 2007 8. In a chemistry lab a student conducts the Hall experiment on an ionic
solution that has equal numbers of positive and negative charge carriers. If the current moves to the right and the magnetic ﬁeld points out of the
page, what will the student measure for the hall voltage AV 2 Vtop — Vbottom?
(a) AV > 0 ,.~ 1 N V
®)AV<O Cf VdL ()AV=O 9. In the three circuits below, all of the batteries are identical.
char e M“ chare.
wmm —%[email protected]{"
H
~er H I II III 5 V T If each circuit has been left in the charging position for a very long I w
time, in which circuit does the capacitor have the most charge on it? Q r Va
(a) I (d) I and II are equal and greater than III , resistors, and capacitors char’ e discharge l
M ‘i‘ “ (b) II (e) I and III are equal and greater than II All have the (c) III (f) II and III are equal and greater than I same charge. When the switch is thrown to the discharging position in each circuit7 in
which circuit will the capacitor discharge fastest? (a) I (d) I and II are equal and faster than III (b) II (e) I and III are equal and faster than II (g) All are equally
III (f) II and III are equal and faster than I fast. 5>K >T PHYS 126 Exam #3 19 April 2007 10. The ﬁgure below shows the cross—section of a loop of wire and the magnetic
ﬁeld that it creates. N Of the two choices below, which best describes the ﬂow of current in the
loop that creates the above ﬁeld? (a) I 11 a’
y. PHYS 126
Exam #3 19 April 2007 Problems (60 points) You must work problem #1, but you may choose to work either problem #2 or
problem #3. Please circle the number of the problem you want graded
on the front ofthe exam. If you do not, you will be penalized 2 points on the problem. 1. (30 points) This problem continues on the next page. An antiproton
(same properties as a proton except that q 2 —e) is moving upward with a
speed of 400 m/s in an electric ﬁeld that points to the left with a strength
of 1700 N/C and a magnetic ﬁeld that points into the plane of the page
with a strength of 425 T. x y: ‘1 {figs} C
M, —\ 4093‘” f 6 7 4 Q37 am J: t) r  'F (It) 9’;
7:...) be __ ( , ” + \f
x \ (a) What acceleration (magnitude and direction) does the antiproton feel
at this instant? I
I X PHYS 126 Exam #3 19 April 2007 (b) What acceleration (magnitude and direction) would the antiproton
feel if it were moving downward with the same speed? ’ , o
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\/ K K ’ ., ('q
:7 ’6‘. vk{o,\ﬁc)( (1/00 %(“\\(p)(\\U 47 76,44 X'OI'VN PHYS 126
Exam #3 19 April 2007 2. (30 points) This problem continues on the next page. A 50—pF
capacitor is wired into the circuit below with two 2.004(5) resistors and a
9—v battery that has no internal resistance. (a) (5 points) If the switch has been left open for a very long time, what
is the potential across the capacitor the instant the switch is closed? \jz’ﬁ,
i ELOV\'/ (b) (5 points) After the switch has been closed a very long time, what is
the potential across the capacitor? ' qu/ZZVSV ,
g s PHYS 126 Exam #3 19 April 2007 (c) (10 points) Assuming the switch has been closed for a very long time and is then suddenly opened, how long does it take for the potential difference across the capacitor to reach half the value found
in part (b)? Q : CAVC " VU T: KC : 3&0ng ©Odewf>i A5
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z/ «1"? QCQXSJSJL> “UQ5X\\§ : /t (d) (10 points) At the instant in part (c), i.e. when the switch has been
left open the amount of time you found in part (c), What is the current
through the topmost resistor and the current through the rightmost VelK gnqiwgw
: O 7620 6* VP .075 x, :L
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Goon); oaoxuﬂg AVL‘Q—tct‘OXW C: Oomrv 2.251/ C 50Mo'b’f f t % :Q‘OWLV :' qu/HO'U4‘ "‘ 2 gnaw 10 PHYS 126 Exam #3 19 April 2007 3. (30 points) This problem continues on the next page. The outer
coil of wire is 10.00 cm long, 2.00 cm in diameter, wrapped tightly with
15 turns of wire, and has a total resistance of 1.0 Q. It is attached to a
battery, as shown, that steadily decreases in voltage from 12 V to 0 V in
0.5 s, then remains at 0 V for t > 0.5 s. The inner coil of wire is 1.00 cm long, 1.00 cm in diameter, has 3 turns of
wire, and has a total resistance of 0.01 9. It is connected, as shown, to a
current meter that has negligible resistance. V (a) (10 points) What is the initial (t : O) magnitude and direction of
the magnetic ﬁeld inside the solenoid? ,) [34' \I’: \l
i : T~ if 713.50 l" \‘K
3 [A’OMLM‘Q’X' J xmw A .4/2’
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T A, /ri \, PHYS 126 Exam #3 19 April 2007 (b) (10 points) What current is induced in the meter during the ﬁrst 0.5 s? In which direction (@~to—right or right—toleft) does this cur—
rent ﬂow through the meter? i: it; \7] k ,r
(C : Ca I v, F (%\\<D:>
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iv 31‘; :33) Mi/ ‘\ \i x k fit ‘(xJOwI (f : gawo’f‘wm : 5‘ 3J’MQ4/M‘;
: E‘OQD/ :o’
:— i‘OWX‘KQ/év : \ULQWO“ “Bi/1C
O i JL, ' (c) (10 points) What current is induced in the meter after the ﬁrst 0.5 s? In which direction (lefttowight or right—to—Ieft) does this current ﬂow
through the meter? ".5 ~g’
Wow/m}; mad/vi W95: 9‘ iii/W (a a:
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tm+—wmteMWwwf 12 Exam #3 Name: W Exercise 1 10 PHYS 126
19 April 2007 ID: S0 LO‘l QQBV Steps Wire and loop (c) No induced current when steady I
(a) Force is up on loop (a) AV21 = +10 V (d) Direction of current is 2, then 1
Force points down (b) gold Wire maximizes B (a) all charge ﬂows through wire
I will also accept (d) none of the above e—e'l' production
(b) positron is dashed path 2
(a) fastest is solid path 1 (c) No potential difference can be
created RC circuits (g) All capacitors have the same charge
(0) circuit III has the smallest time con~
stant (b) Current moves cw as seen from
above Total 4
2
2
4‘
4. 4 40 Max Score ﬁ:
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4/
2
7/ LL:
gt Exam #3 Problem Steps 1 Antiproton in crossed ﬁelds Magnitude of magnetic force Students must use FB = [(117 X Students must use [q] = e = 1.6 x 10"19 C,
v = 400 m/s, and B = 4.25 T Students should ﬁnd that magnitude of mag—
netic force in both problems is F3 2 evB is 2.72 x 10"16 Magnitude of electric force Students must use = lqu Students must use IqI = e = 1.6 x 10—19 C,
and E 21700 N/C Students should ﬁnd that magnitude of elec
tric force in both problems is F}; = eE z
2.72 x 10*“5 Magnitude of accelerations One of the student’s answers should be due
to the forces adding because they point in the
same direction One of the student’s answers should be due
to the forces subtracting because they point
in opposite directions Students must use Newton’s second law to
find accelerations in both cases a = [Fnetl/m Students must mass of proton m = 1.67 x
10‘27 kg Students should ﬁnd that the magnitude of
the accelerations are a = IFE i FBI/m z
0 or 3.26 x1011 m/s2 Directions of accelerations The acceleration direction for part (a) is to
the right The acceleration direction for part (b) is 5 Total PHYS 126
19 April 2007 Max Score
30
6 2 .2
2 34
:1: 6
2 3;
2 j:
2 l
10 Iv INN" ii
48c
we: PHYS 126 Exam #3 19 April 2007
Problem Steps Max Score
2 RC circuit 30
2a IAVCI when is just closed 5 AVG = 0 5 ;
2b IAVCI when is closed for long time 5 Correct: IAVCI = 4.5 V 5 Partial: lAVc = 9 V 2 z /
2c Find halflife 10 Find time constant using ’1’ = R0 (correct 3 answer is 0.1 s) Use0=50pF=5X10“5F 2 UseR=2ko=2x103o 2 Use V = Voe“t/T to ﬁnd that time for half 2 of the potential to deay away is t = Tln2 z 0.70 X 7’ (correct answer is 0.069 5) Correct units 1
2d Currents 10 Topmost resistor 7 Use I = Efﬁe—VT or Ohm’s law to ﬁnd 2 current I = AV/R (Correct answer is I a: 1.1 x 10"”3 A) Use potential across capacitor from part (c), 2 X i.e. half of the potential quoted in part (b). (Correct potential is 2.25 V) UseR=2kQ=2x103Q 2 Correct units 1 Rightmost resistor 3 Current is zero. 3 Total 30 Exam #3 Problem Steps 3 Induction coils 3a Find magnetic ﬁeld
Find current
Use Ohm’s law to ﬁnd current I = AV/R
(correct answer is 12 A)
Use voltage of 12 V and a. resistance of 1.0 9
Find ﬁeld strength
Use Bi0t~Savart law to ﬁnd ﬁeld B 2 “0%1
(correct answer is 2.26 X 10"3 T)
UseN=15andL=0.1m
Correct units
Direction is to the right 3b Find induced current 0 g t S 0.5 3
Find area using A = N§D2 where N = 3
and D = 0.01 m. (Correct answer is A z
2.4 ><10“4 m2.)
Find rate of change of Bﬁeld using AX? where
AB is just the answer from part (3.) because
the ﬁnal ﬁeld is zero, and At = 0.5 s. (C0r~
rect answer is 95 R: 4.5 X 10—3 T/si)
Find EMF using rate of change of ﬂux %? =
[4%. (Correct answer is 8 z 10‘6 V.)
Find induced current using Ohm’s law I :
S/R. (Correct answer is I w 10—4 A.)
Correct units
Direction is Right to Left in meter 3c Find induced current t > 0.5 s No current is induced Total Max
30 10 (04k {003 10
10 3O PHYS 126
19 April 2007 Score ...
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 Spring '08
 MUTCHLER
 Magnetic Field, Electric charge

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